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(15.2.3)   Let P(X) = a0 4- oqX H ----- h anXn be a nonconstant polynomial
with complex coefficients, with ocn ^ 0. For each x e A, put

P(x) = a0 e 4- #1* + • • • + anxn e A.
For each complex number /^, we can write P(X) — n in the form

where the Aj are complex numbers (9.11.2). Since the subalgebra of A genera-
ted by e and x is commutative, we have

P(x) - jw? = an(x - V) "•(*- AB e).

Hence P(x) — \JLQ is invertible in A if and only if all the x — ^e are invertible
in A; consequently \JL e SpA(P(Xŧ if and only if lj e SpA(x) for at least one
value of y, and since JLI = P(/lj) for each j, this condition is equivalent to
\i e P(SpA(x)). Therefore

(                              SpA(P(x)) = P(SpA(x)).

Suppose now that 0 ^ P(SpA(;c)), so that P(*) is invertible in A. Let Q(X)
be another polynomial with complex coefficients, and put

R(X) = Q(X)/P(X).

Then we may form the element Q(x)(P(;c))~1 = (P(x))~lQ(x) of A, which we
denote by R(x). Then we have

(                              SpA(R(*)) = R(SpA(*)).

For, replacing R by R — ^u for some /i e C if necessary, we have to show
that R(x) is invertible if and only if 0 ^ R(SpA(x)); or equivaiently that Q(x)
is invertible if and only if 0 Ģ Q(SpA(X)). But this follows from (15.12.3) if
Q is not a constant, and it is obvious if Q is a constant.

(1 5.2.4)   Let A be a Banach algebra with unit element e^Q.

(i) The multiplicative group G of invertible elements of A is open in A,
and contains the ball \\x — e\\ < 1. The topology induced on G by that of A. is
compatible with the group structure of G.

(ii) For each x e A, the set Rx = C — SpA(x) of regular values, for x is
open in C, and the mapping C i— > (x — (e)~ * of Rx into A is analytic.

(iii)   The set SpA(x) is compact and nonempty, and is contained in the ball

ici ^ n*ii-  Every C e C such that x — Ģe is a (left or right) zero-divisor in A is a